why can′t we differentiate or intergrate powers of trigonometric functions such as 1) ∫cos^2 xdx? 3) ∫tan^2 2xdx 2)∫sin^2 xdx? 4) ∫sin^(10) x hence how do we solve such problems.?


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 64017 by Rio Michael last updated on 12/Jul/19

$w h y {c a n}^{'} t w e d i f f e r e n t i a t e o r i n t e r g r a t e p o w e r s o f t r i g o n o m e t r i c$ $f u n c t i o n s s u c h a s$ $1) \int {c o s}^{2} x d x ? 3) \int {t a n}^{2} 2 x d x$ $2) \int {s i n}^{2} x d x ? 4) \int {s i n}^{10} x$ $h e n c e h o w d o w e s o l v e s u c h p r o b l e m s . ?$
Commented by kaivan.ahmadi last updated on 12/Jul/19

$c o s 2 x = 2 {c o s}^{2} x - 1 \Rightarrow {c o s}^{2} x = \frac{1 + c o s 2 x}{2}$ $\int {c o s}^{2} x d x = \int \frac{1 + c o s 2 x}{2} d x = \frac{1}{2} \int (1 + c o s 2 x) d x =$ $\frac{1}{2} (x + \frac{1}{2} s i n 2 x) + C$
Commented by kaivan.ahmadi last updated on 12/Jul/19

$c o s 2 x = 1 - 2 {s i n}^{2} x \Rightarrow {s i n}^{2} x = \frac{1 - c o s 2 x}{2}$ $\int {s i n}^{2} x d x = \frac{1}{2} \int (1 - c o s 2 x) d x =$ $\frac{1}{2} (x - \frac{1}{2} s i n 2 x) + C$
Commented by kaivan.ahmadi last updated on 12/Jul/19

$\int {t a n}^{2} 2 x d x = \int (1 + {t a n}^{2} 2 x - 1) d x = \frac{1}{2} t a n 2 x - x + C$
Commented by MJS last updated on 12/Jul/19

$look at qu . 64037$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum